Question

A block of mass 23 Kg rests on a plane making an
angle of 100 with horizontal. Determine the
component of the weight normal to the plane. Select
one:​

Answer 1

Given:-

Mass of block, m=23kg

Angle with horizontal=$100^{0}$

To Find:-

Component of weight normal to the plane

Explanation:-

Since weight(mg) will be vertically downward and this mass is inclined to $10^{0}$ with the horizontal. Now,

Component of force along the plane, $\text{F_{s}}=\text{mgcos10^{0}}$$F_{s}=mgcos10^{0}$

Component of force normal to the plane, $F_{N}=mgsin10^{0}$

Substitute values of m, g and sin$10^{0}$ in above equation.

$F_{N}=23\times9.81\times sin10^{0}\\\\F_{N}=23\times9.81\times0.1736=39.18N$